Question

An aqueous KCl solution of density 1.20 g mL^–1 has a molality of 3.30 mol kg^–1 . The molarity of the solution in mol L^–1 is ........... (Nearest integer) [Molar mass of KCl = 74.5]

Answer 1

Answer is 3.17

1000 kg solvent has 3.3 moles of KCl 

1000 kg solvent → 3.3 x 74.5 gm KCl 

→ 245.85

Weight of solution = 1245.85 gm

Volume of solution = \(\frac{1245.85}{1.2}\) = 1.0382 L

NOW, MOLARITY=moles of KCl / Vol. of Sol. in lt = 3.3/1.0382=3.17 mol/L

Answer 2

3.3 molal =3.3 moles of KCl in 1Kg of Water {Since solution is aqueous}

Wt of KCl in gms=moles of KCl x Molar Mass of KCl=74.5x3.3=245.85 gms

Wt of Solution in gms=Wt of Solute(in gms)(i.e KCl)+Wt of solvent(in gms)(i.e Water)

Wt of Solution = 1245.85gms

Volume of Solution=Wt of Solution/Density of Solution=1245.85/1.2=1038.20mL=1.0382L

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NOW, MOLARITY=moles of KCl / Vol. of Sol. in lt = 3.3/1.0382=3.17 mol/L(ans)